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I am considering the Schwarzschild metric. I have calculated my Christoffel symbols and am able to calculate the Riemann tensor (I think). In short, I have done a bunch of work to find this thing called the Riemann curvature tensor. As I understand it, if I take a vector $Z$ at a point on my manifold, $R(X,Y)Z$ gives me the difference between my original vector and the parallel transported one. But I really don't know what to do next. I want to use this tensor to compute some useful piece of physical information. For instance, I can use $$v_f = v_i +at$$ to calculate the final velocity of an object dropped for a duration $t$ and an acceleration $a = g$. This is an example of an application that yields interesting/useful information. I want something that allows me to do something similarly interesting with my curvature tensor.

My Question:

Can someone provide me with an example of what I can do with my curvature tensor? What's an example of a useful application?

EDIT: On page 118 of Wald, he says

However, the Schwarzschild solution, which describes the exact exterior field of spherical body, predicts tiny departures from Newtonian theory for the motion of planets in our solar system, and, in addition, predict the "bending of light", the gravitational redshift of light, and "time delay" effects. These four predictions have been accurately confirmed by precise measurements.

Thus, I assume then that one of those four examples must require use of the Riemann tensor somehow. Perhaps someone can explain which one(s) do and how I might go about using my curvature tensor to describe these phenomena.

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  • $\begingroup$ From memory, try the schwardschild solution first, less variables, just r and t, less drudgery if you can hold the angular variables fixed. It's still uses the Einstein equation but leaves out non essential bits $\endgroup$
    – user74893
    Commented Mar 28, 2015 at 19:51
  • $\begingroup$ To expand a bit, its the Einstein equation with r.h.s set to zero , empty space, and cosmological constant left out. Again, this is from memory, but I found it easiest to follow through on the implications of curved space. Apologies if I have misunderstood your question $\endgroup$
    – user74893
    Commented Mar 28, 2015 at 19:54
  • $\begingroup$ Why do we set the Stress-Energy Tensor equal to zero? $\endgroup$
    – Stan Shunpike
    Commented Mar 28, 2015 at 19:59
  • $\begingroup$ Or rather, why are we considering empty space? I thought the point was to consider something with gravity influencing it... $\endgroup$
    – Stan Shunpike
    Commented Mar 28, 2015 at 20:03
  • $\begingroup$ Outside black hole there is assumed to be no energy or mass, so it's homogenous equation. Less variables to worry about, then try the kerr solution for rotating b. h. You will see what I mean when you try it :) $\endgroup$
    – user74893
    Commented Mar 28, 2015 at 20:03

2 Answers 2

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One thing you can use the curvature tensor for is to detect singularities in the spacetime. For the Schwarzschild solution, the simpler curvature scalars formed from the Ricci tensor, $R$ and $R_{ab} R^{ab}$ vanish everywhere due to the fact that this is a vacuum solution to the Einstein equations. But since you have the full Riemann tensor at your disposal, you can compute what is called the Kretschmann scalar, $R_{abcd} R^{abcd}$. You should find that at $r = 2M$ this scalar is finite, whereas the Schwarzschild metric in the usual coordinates is singular there. This is because the singularity at $r=2M$ in the metric is just due to a bad choice of coordinates. However, if you evaluate this scalar at $r=0$, you will find that it diverges to infinity, indicating a curvature singularity. This is a true singularity that is cannot be done away with by a change of coordinates.

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  • $\begingroup$ @0celo7 True, in general using the curvature scalars you aren't guaranteed to find singularities. But it provides a useful sufficient condition that applies in a large variety of cases. $\endgroup$
    – asperanz
    Commented Mar 31, 2015 at 18:27
  • $\begingroup$ @0celo7 Also FRW universes do have a curvature singularity at the Big Bang. The Hubble rate contributes to the spacetime curvature and that diverges at the Big Bang. $\endgroup$
    – asperanz
    Commented Apr 1, 2015 at 22:24
  • $\begingroup$ Ah yes, it is the spatial curvature which is constant. My apologies. $\endgroup$
    – Ryan Unger
    Commented Apr 1, 2015 at 22:59
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You have the order wrong in which you "get" things. You get the energy-momentum tensor from your specific matter theory. You do not know what $g_{ab}$ is. Then, given some general assumptions about your spacetime, you write down an ansatz for the metric. For Schwarzschild, we have time independence and $\mathrm{SO}(3)$ isometry. Then you calculate the Ricci tensor and scalar curvature of your ansatz and plug it into the Einstein equations, with the energy momentum tensor obtained above on the RHS. Then you solve the Einstein equations$^{(1)}$. Integration constants are determined by various methods.

All of those problems use the metric itself in the form of the geodesic equation, not the Riemann tensor.

$^{(1)}$This is about as simple as "killing the Batman".

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